Answer
930 V
3.9 A
Work Step by Step
First use the relationship between peak and rms values, equation 18–8.
$$V_{peak}=\sqrt2V_{rms}=\sqrt{2}(660V)=930V$$
Next, relate the rms values to the power by using equation 18-9a.
$$P=I_{rms}V_{rms}=\frac{I_{peak}}{\sqrt2}V_{rms}$$
$$I_{peak}=\frac{\sqrt2P}{V_{rms}}=\frac{\sqrt2(1800W)}{660V}=3.9A$$
